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Chapter 11: Problem 6

For the reaction $$5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}$$ it was found that at a particular instant bromine was being formed at the rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). At that instant, at what rate is (a) water being formed? (b) bromide ion being oxidized? (c) \(\mathrm{H}^{+}\) being consumed?

Short Answer

Expert verified
Question: At a particular moment in a reaction, the rate of bromine formation is 0.039 mol/L⋅s. Determine the rates of formation or consumption of water, bromide ion, and hydrogen ion at that instant. Answer: At that particular instant, water is being formed at a rate of 0.039 mol/L⋅s, bromide ions are being oxidized at a rate of 0.065 mol/L⋅s, and hydrogen ions are being consumed at a rate of 0.078 mol/L⋅s.

Step by step solution

01

a) Rate of water formation

To determine the rate of water formation, we need to compare the stoichiometric coefficients of water and bromine in the balanced equation. We can see that the coefficients are 3 and 3, respectively. Since the rate of bromine formation has been given as \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), the rate of water formation would be the same since their coefficients are equal. Therefore, the rate of water formation is: $$0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$
02

b) Rate of bromide ion oxidation

To find the rate at which bromide ions are being oxidized, we need to compare the stoichiometric coefficients of bromide ions and bromine in the balanced equation. We can see that they are 5 and 3, respectively. To find the rate of bromide ion oxidation, we can set up a proportion: $$\frac{\text{Rate of Br}^{-}}{\text{Rate of Br}_{2}} = \frac{5}{3}$$ Therefore, the rate of bromide ion oxidation is: $$\text{Rate of Br}^{-} = \frac{5}{3} \times 0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} = 0.065 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$
03

c) Rate of H+ ion consumption

To determine the rate at which H+ ions are being consumed, we need to compare the stoichiometric coefficients of H+ ions and bromine in the balanced equation. We can see that they are 6 and 3, respectively. To find the rate of H+ ion consumption, we can set up a proportion: $$\frac{\text{Rate of H}^{+}}{\text{Rate of Br}_{2}} = \frac{6}{3}$$ Therefore, the rate of H+ ion consumption is: $$\text{Rate of H}^{+} = 2 \times 0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} = 0.078 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}.$$ In summary, at that particular instant, water is being formed at a rate of \(0.039 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), bromide ions are being oxidized at a rate of \(0.065 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), and H+ ions are being consumed at a rate of \(0.078 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\).

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The reaction $$\mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{NOBr}(g)$$ is second-order in nitrogen oxide and first-order in bromine. The rate of the reaction is \(1.6 \times 10^{-8} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min}\) when the nitrogen oxide concentration is \(0.020 \mathrm{M}\) and the bromine concentration is \(0.030 \mathrm{M}\). (a) What is the value of \(k\) ? (b) At what concentration of bromine is the rate \(3.5 \times 10^{-7} \mathrm{~mol} / \mathrm{L} \cdot \min\) and \([\mathrm{NO}]=0.043 \mathrm{M} ?\) (c) At what concentration of nitrogen oxide is the rate \(2.0 \times\) \(10^{-6} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min}\) and the bromine concentration one fourth of the nitrogen oxide concentration?
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